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Standard XII

Physics

Forced Oscillations

Due to some f...

Question

Due to some force $F_{1}$ a body oscillates with period ( $4 / 5$ ) $s$ and the due to other force $F_{2}$ it oscillates with period $3 / 5 s$ . if both the forces acts simultaneously new period will be

0.72 s

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0.64 s

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0.48 s

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0.36 s

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Solution

The correct option is C $0.48 s$
$F = \frac{4 π^{2} m x}{T^{2}}$
$F_{1} = \frac{4 π^{2} m x}{T^{2}}$
$F_{2} = \frac{4 π^{2} m x}{T_{2}^{2}}$
$F_{1} + F_{2} = \frac{4 π m x}{T^{2}}$
$∴ \frac{1}{T^{2}} = \frac{1}{T_{1}^{2}} + \frac{1}{T_{2}^{2}}$
$\frac{1}{T^{2}} = (\frac{5}{4})^{2} + (\frac{5}{3})^{2}$
$T = 0.48$

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Due to some force F1 a body oscillates with period (4/5) s and the due to other force F2 it oscillates with period 3/5s. if both the forces acts simultaneously new period will be

The correct option is C 0.48sF=4π2mxT2F1=4π2mxT2F2=4π2mxT22F1+F2=4πmxT2∴1T2=1T21+1T221T2=(54)2+(53)2T=0.48

Due to some force $F_{1}$ a body oscillates with period ( $4 / 5$ ) $s$ and the due to other force $F_{2}$ it oscillates with period $3 / 5 s$ . if both the forces acts simultaneously new period will be